1 Invariant subspaces


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1 MATH 2040 Linear Algebra II Lecture Notes by Martin Li Lecture 8 Eigenvalues, eigenvectors and invariant subspaces 1 In previous lectures we have studied linear maps T : V W from a vector space V to another vector space W. As we have seen before, an important special case is when V = W. Such a linear map T L(V ) is called an operator on V. We will study linear operators in this lecture and the next. From now on, V denotes a vector space over F and I L(V ) is the identity map on V. 1 Invariant subspaces Our major goal is to understand what a linear operator T L(V ) looks like. Suppose we have a direct sum decomposition V = U 1 U 2 U m into subspaces U 1,, U m of V. By linearity, to understand the operator T we only need to know the behaviour of T on each subspace U i, which potentially is easier as U i has smaller dimension than V (provided they are finite dimensional). However, to carry out this reduction process, we need to make sure that the image of U i under the operator T to be contained inside the subspace U i itself. Only then it is possible for us to regard T as an operator on the subspace U i. This motivates the definition below. Definition 1. Let T L(V ). A subspace U of V is said to be invariant under T if T (U) U, i.e. T u U for all u U. In this case, the restriction T U : U U is a linear operator on U, i.e. T U L(U). Sometime we also say U is a T invariant subspace. We first look at some examples of invariant subspaces. Example 1. Let V be a vector space over F. The subspaces {0} and V are invariant under T for any T L(V ) ( why?). Example 2. Let T L(V ). The subspaces ker T and range T are invariant under T. To see this, suppose v ker T, then T v = 0 ker T. Therefore T (ker T ) = {0} ker T and ker T is invariant under T by definition. On the other hand, suppose v range T, then v = T u for some u V. Therefore, T v = T (T u) range T and so range T is invariant under T. Do we have other examples different from the ones above? Recall that our motivation of defining invariant subspaces U is to reduce the study of T to a smaller subspace. It does not really help much if U = {0} or V. The invariant subspaces in Example 2 may not help much either as they could be {0} 1 last revised on October 16,
2 or V (consider for example the zero transformation and the identity map). Later, we will see that one can in fact construct nontrivial invariant subspaces for any linear operators on a finite dimensional vector space V, provided that dim V > 1 when F = C or dim V > 2 when F = R. Example 3. Let T : R 3 R 3 be the linear operator defined by T (x, y, z) = (x + y, y + z, 0). Then the xyplane and the xaxis are invariant subspaces under T. Exercise 1. Let T : R 4 R 4 be the linear operator defined by T (x 1, x 2, x 3, x 4 ) = (x 1 + x 2 + 2x 3 x 4, x 2 + x 4, 2x 3 x 4, x 3 + x 4 ). Show that the x 1 x 2 plane is an invariant subspace under T. Example 4. Consider the differentiation map D : P(R) P(R), which is a linear operator on the infinite dimensional vector space P(R), the subspace P m (R) is invariant under D for any nonnegative integer m. Therefore, we can restrict D Pm(R) : P m (R) P m (R) to a finite dimensional subspace (this is a useful trick that has been used several times so that some results, e.g. the Fundamental Theorem of Linear Algebra, requiring the finite dimensionality of the vector spaces can be applied). The following example shows that it may not always be possible to decompose V into a direct sum of smaller invariant subspaces. Example 5. Let T : F 2 F 2 be the linear operator defined by T (x, y) = (y, 0). Then, U = {(x, 0) : x F} is an invariant subspace under T. In fact, T (U) = {0} U. However, there does not exist another subspace W F 2 which is invariant under T such that F 2 = U W. To see this, suppose such an invariant subspace W exists, we must have dim W = 1. Therefore, W = span{w} where w F 2 is an eigenvector of T. However, it is easy to see (why?) that 0 is the only eigenvalue of T and all eigenvectors of T are in U. This gives a contradiction to the assumption that F 2 = U W and therefore such a W cannot exist. Exercise 2. Let T L(V ) and U be a subspace of V. Prove that (a) if U ker T, then U is invariant under T. (b) if range T U, then U is invariant under T. Exercise 3. Suppose S, T L(V ) such that ST = T S. Prove that ker S and range S are both invariant under T. Exercise 4. Let T LL(V ) and U 1,, U m be invariant subspaces of V under T. Show that U U m and U 1 U m are both invariant under T. *Exercise 5. Prove or give a counterexample: If U is a subspace of V that is invariant under every operator T L(V ), then U = {0} or U = V. 2
3 2 Eigenvalues and eigenvectors We now turn to the study of the simplest invariant subspaces which is nontrivial. These are invariant subspaces of dimension 1. Let T L(V ) and U be an invariant subspace of V under T. Suppose dim U = 1. Then there exists a basis {v} of U such that v 0. Since span{v} = U and T (U) U, we must have T v = λv for some λ F. Conversely, if v V is a nonzero vector such that T v = λv for some λ F. Then, U = span{v} is a 1dimensional invariant subspace of V under T. Therefore, we have a complete description of 1dimensional invariant subspaces (if exist). Definition 2. Let T L(V ). A scalar λ F is called an eigenvalue of T if there exists a nonzero vector v V such that T v = λv. Such a nonzero vector v V is said to be an eigenvector of T corresponding to λ. Remark 1. In some textbooks, the terms characteristic value and characteristic vector are used instead of eigenvalues and eigenvectors. Note that each eigenvector has a unique eigenvalue associated to it but an eigenvalue can have different (even linearly independent) eigenvectors. Remark 2. By definition, an eigenvector v has to be a nonzero vector since T 0 = λ0 for any λ F. However, it is possible to have λ = 0 as an eigenvalue. Example 6. Let T 0 L(V ) be the zero transformation, i.e. T 0 v = 0 for all v V. nonzero v V is an eigenvector of T 0 with eigenvalue 0. Then every Example 7. Let I V L(V ) be the identity map on V, i.e. I V v = v for all v V. Then every nonzero v V is an eigenvector of I V with eigenvalue 1. The following example shows that the field F plays a role on the existence of eigenvalues/eigenvectors. Example 8. Let T : F 2 F 2 be the linear operator defined by T (w, z) = ( z, w). We want to find all the eigenvalues and eigenvectors of T. Suppose (w, z) F 2 is an eigenvector of T with eigenvalue λ F. By definition (w, z) (0, 0) and T (w, z) = λ(w, z). Using the definition of T and equating each component, we have Combining these two equations we have z = λw and w = λz. z = λ 2 z and w = λ 2 w. 3
4 As (w, z) (0, 0), at least one of them, w or z, is nonzero. Therefore, we must have λ 2 = 1. If F = R, there is no such λ R so T has no eigenvalues or eigenvectors when F = R. When F = C, we conclude that λ = i or i. In fact, the eigenvectors corresponding to the eigenvalue i are vectors of the form (w, iw) with 0 w C, and the eigenvectors corresponding to the eigenvalue i are vectors of the form (w, iw) with 0 w C. Exercise 6. Give an example of a linear operator T : R 4 R 4 such that T has no real eigenvalues. Exercise 7. Let T : R 2 R 2 be the linear operator defined by T (x, y) = ( 3y, x). eigenvalues and eigenvectors of T. Find all the Exercise 8. Let T : F 2 F 2 be the linear operator defined by T (w, z) = (z, w). Find all the eigenvalues and eigenvectors of T when F = R and F = C. Exercise 9. Let T : F 3 F 3 be the linear operator defined by T (z 1, z 2, z 3 ) = (2z 2, 0, 5z 3 ). Find all the eigenvalues and eigenvectors of T when F = R and F = C. Exercise 10. Let T : F n F n be the linear operator defined by T (z 1, z 2, z 3,, z n ) = (z 1, 2z 2, 3z 3,, nz n ). Find all the eigenvalues and eigenvectors of T when F = R and F = C. Can you describe all invariant subspaces of T. Proposition 3. Let T L(V ) where V is a vector space over F. Then λ F is an eigenvalue of T T λi is not injective Proof. Note that the equation T v = λv is equivalent to (T λi)v = 0. eigenvalue if and only if ker(t λi) {0}, i.e. T λi is not injective. Therefore, λ F is an Remark 3. When V is finite dimensional, injectivity of an operator on V is equivalent to surjectivity (recall Proposition 5 in Lecture 7). In this case, we have that λ is an eigenvalue of T if and only if T λi is not invertible. The latter condition is in turn equivalent to the noninvertibility of the matrix [T λi] β with respect to any basis β of V. This gives the standard method of finding eigenvalues of a matrix by solving the characteristic polynomial equation. Exercise 11. Let T L(V ) where V is a complex finite dimensional vector space (i.e. F = C). Suppose that there exists some basis β of V such that [T ] β M n n (R) where n = dim V. Show that λ C is an eigenvalue of T if and only if λ C is an eigenvalue of T. Exercise 12. Prove that a linear operator T : R 3 R 3 must have at least one real eigenvalue. The proof of Proposition 3 also gives a characterization of the eigenvectors of T. Proposition 4. Let T L(V ) and λ F be an eigenvalue of T. Then v V is an eigenvector of T corresponding to λ 0 v ker(t λi) 4
5 Example 9. Let D : P(R) P(R) be the linear operator Dp = p. We want to find all the eigenvalues and eigenvectors of D. Suppose p P(R) is an eigenvector of D with eigenvalue λ R. Then, p = Dp = λp. If λ 0, then deg p = deg(λp) = deg p, which is impossible since deg p = deg p 1 (note that p is a nonzero polynomial as it is an eigenvector). Therefore, we must have λ = 0 and thus p = 0, which means p(x) = a 0 for some nonzero a 0 R. Therefore, the only eigenvalue of D is 0 with eigenvectors being all the nonzero constant polynomials. One can also compute the eigenvalues and eigenvectors using matrices. However, P(R) is infinite dimensional so we restrict the operator D : P m (R) P m (R) to the finite dimensional invariant subspace P m (R) (recall Example 4). Using the standard basis {1, x,, x m } of P m (R), the matrix of D with respect to β is given by (recall Example 14 in Lecture 5) the following (m + 1) (m + 1) matrix: [D] β = m Therefore, the characteristic polynomial equation is given by det([d] β λi) = ( λ) m+1 = 0. Therefore, λ = 0 is the only eigenvalue of D. To find the eigenvectors of D corresponding to the eigenvalue 0, 1 0 note that the kernel of the matrix [D] β is equal to span, which corresponds to the eigenvectors. 0 of the form p(x) = a 0 0 under the basis β. Exercise 13. Let T : P 4 (R) P 4 (R) be the linear operator defined by T (p(x)) = xp (x). Find all the eigenvalues and eigenvectors of T. Exercise 14. Let T : M n n (R) M n n (R) be the linear operator defined by T (A) = A t. Find all the eigenvalues and eigenvectors of T. *Exercise 15. Let T : F F be the backward shift linear operator defined by Find all the eigenvalues and eigenvectors of T. T (z 1, z 2, z 3, ) = (z 2, z 3, ). Exercise 16. Let T, S L(V ) where S is invertible. Prove that T and S 1 T S have the same eigenvalues. What is the relationship between their eigenvectors? Exercise 17. Let T L(V ) be invertible. (a) Show that 0 is not an eigenvalue of T. (b) Prove that λ F is an eigenvalue of T if and only if λ 1 F is an eigenvalue of T 1. (c) Prove that T and T 1 have the same eigenvectors. 5
6 Exercise 18. Let S, T L(V ) where V is finite dimensional. Prove that ST and T S have the same eigenvalue. Now, we establish some general facts about eigenvalues and eigenvectors. Proposition 5. Let T L(V ). Suppose λ 1,, λ m F are distinct eigenvalues of T with corresponding eigenvectors v 1,, v m V. Then {v 1,, v m } is linearly independent. Proof. We argue by contradiction. Suppose that the subset {v 1,, v m } is linearly dependent. Then there exists a smallest positive integer k such that v k span{v 1,, v k 1 }. Therefore, there exists a 1,, a k 1 F such that v k = a 1 v a k 1 v k 1. Applying the operator T on both sides and using that T v i = λ i v i, we have λ k v k = a 1 λ 1 v a k 1 λ k 1 v k 1. Note that some a i is nonzero, otherwise v k = 0 which cannot be an eigenvector. Multiplying the first equation by λ k and then subtract from the second equation, we obtain 0 = a 1 (λ k λ 1 )v a k 1 (λ k λ k 1 )v k 1. As the eigenvalues λ i are distinct, this means that the subset {v 1,, v k 1 } is linearly dependent and therefore there exists some positive integer j k 1 such that v j span{v 1,, v j 1 }. This contradicts the choice of k to be the smallest of such positive integer. Therefore, we must have {v 1,, v m } is linearly independent. Exercise 19. Let T L(V ) where V is finite dimensional. Suppose W is an invariant subspace of V under T. Prove that if v 1,, v k V are eigenvectors of T corresponding to distinct eigenvalues λ 1,, λ k F such that v v k W, then v i W for all i = 1,, k. (Hint: Use induction on k.) *Exercise 20. Let V be a finite dimensional vector space over F and v 1,, v m V. Prove that {v 1,, v m } is linearly independent if and only if there exists T L(V ) such thatv 1,, v m are eigenvectors of T corresponding to distinct eigenvalues of T. The proposition above have the following consequence. Corollary 6. Let T L(V ). Suppose V is finite dimensional with dim V = n. Then there can be at most n distinct eigenvalues of T. Proof. This follows from Proposition 5 and the fact that any subset of m vectors in an ndimensional vector space must be linearly dependent if m > n. Exercise 21. Let T : R 3 R 3 be a linear operator and 1, 2, 3 are eigenvalues of T. Prove that there exists v R 3 such that T v 5v = (1, 2, 3). Exercise 22. Let T L(V ) where V is finite dimensional. Suppose λ F is some fixed scalar. Prove that there exists α F such that α λ < and T αi is invertible. 6
7 Exercise 23. Suppose V = U W where U and W are nonzero subspaces of V. Define the projection onto U to be the operator P L(V ) such that P (u + w) = u for any u U and w W. Find all the eigenvalues and eigenvectors of P. Exercise 24. Let T L(V ). Suppose u, v V are eigenvectors of T such that u + v is also an eigenvector of T. Prove that u and v correspond to the same eigenvalue. Exercise 25. Let T L(V ) such that every nonzero vector v V is an eigenvector of T. Prove that T = λi for some λ F. *Exercise 26. Let T L(V ) where V is finite dimensional. Suppose every subspace U V with dim U = dim V 1 is invariant under T. Prove that T = λi for some λ F. *Exercise 27. Let T L(V ) where V is finite dimensional with dim V 3. Suppose every 2 dimensional subspace U V is invariant under T. Prove that T = λi for some λ F. Exercise 28. Let T L(V ). Suppose that dim range T = k. Prove that T has at most k + 1 distinct eigenvalues. 7
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